Exam (elaborations) TEST BANK FOR Gas Dynamics 3rd Edition By James John, Theo Keith (Instructor's Solution Manual)
II N S T R U C T O R ’’ S
S O L
... [Show More] U T II O N S MA N UA L
G A S D Y N A M II C S
James E. A. John, Ph.D.
President
Kettering University
Flint, Michigan
Theo G. Keith, Jr., Ph.D.
Distinguished University Professor
Department of Mechanical, Industrial, and Manufacturing Engineering
The University of Toledo
Toledo, Ohio
T H I R D E D I T I O N
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Table of Contents
Chapter 1 Basic Equations of Compressible Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1
Chapter 2 Wave Propagation in Compressible Media . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .17
Chapter 3 Isentropic Flow of a Perfect Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .32
Chapter 4 Stationary Normal Shock Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .56
Chapter 5 Moving Normal Shock Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .82
Chapter 6 Oblique Shock Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .106
Chapter 7 Prandtl–Meyer Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .129
Chapter 8 Applications Involving Shocks and Expansion Fans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .149
Chapter 9 Flow with Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .169
Chapter 10 Flow with Heat Addition or Heat Loss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .207
Chapter 11 Equations of Motion for Multidimensional Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .236
Chapter 12 Exact Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .250
Chapter 13 Linearized Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .272
Chapter 14 Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .290
Chapter 15 Measurements in Compressible Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .339
This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their
courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web)
will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available
to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to
abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on
these materials.
1
Chapter One
BASIC EQUATIONS
OF COMPRESSIBLE FLOW
Problem 1. – Air is stored in a pressurized tank at a pressure of 120 kPa (gage) and a temperature
of 27°C. The tank volume is 1 m3. Atmospheric pressure is 101 kPa and the local acceleration
of gravity is 9.81 m/s2. (a) Determine the density and weight of the air in the tank, and (b)
determine the density and weight of the air if the tank was located on the Moon where the
acceleration of gravity is one sixth that on the Earth.
g 9.81m/ s R 0.287 kJ / kg K
1m
T 27 273 300 C
P P P 120 101 221 kpa
2
3
abs gage atm
= = ⋅
∀ =
= + = °
= + = + =
a) m3
2.5668 kg
(0.287)(300)
221
RT
ρ = P = =
W = mg = ρ∀g = (2.5668)(1)(9.81) = 25.1801N
b) moon earth 3
m
ρ = ρ = 2.5668 kg
W 4.1967N
6
W 1
g
g
W earth earth
earth
moon
moon = = =
Problem 2. – (a) Show that p/ρ has units of velocity squared. (b) Show that p/ρ has the same
units as h (kJ/kg). (c) Determine the units conversion factor that must be applied to kinetic
energy, V2/2, (m2/s2) in order to add this term to specific enthalpy h (kJ/kg).
Air
2
a)
2
2
2
2
3
2
2 3
V
s
m
N s
1 kg m
kg
N m
kg
m
m
p N
m
, kg
m
p N
≈ = ⎟
⎟⎠
⎞
⎜ ⎜⎝
⎛
⋅
− ⋅
= ⎟
⎟
⎠
⎞
⎜ ⎜
⎝
⎛
⎟ ⎟⎠
⎞
⎜ ⎜⎝
⎛
≈
ρ
⎟ ⎟⎠
⎞
⎜ ⎜⎝
⎛
≈ ρ ⎟
⎟⎠
⎞
⎜ ⎜⎝
⎛
≈
b)
kg
kJ
1000
1
1000 J
1kJ
N m
1 J
kg
P N m = ⎟⎠
⎞
⎜⎝
⎛
⎟⎠
⎞
⎜⎝
⎛
⋅
⋅
≈
ρ
c)
c
2
2
2 2
1000g
factor 1
kg
kJ
1000 J
1 kJ
N m
1 J
kg m
1 N s
s
m
2
V
∴ =
≈ ⎟⎠
⎞
⎜⎝
⎛
⎟⎠
⎞
⎜⎝
⎛
⋅ ⎟
⎟
⎠
⎞
⎜ ⎜
⎝
⎛
⋅
⋅
≈
Problem 3. – Air flows steadily through a circular jet ejector, refer to Figure 1.15. The primary
jet flows through a 10 cm diameter tube with a velocity of 20 m/s. The secondary flow is through
the annular region that surrounds the primary jet. The outer diameter of the annular duct is 30
cm and the velocity entering the annulus is 5 m/s. If the flows at both the inlet and exit are
uniform, determine the exit velocity. Assume the air speeds are small enough so that the flow
may be treated as an incompressible flow, i.e., one in which the density is constant.
m& i = m& e
m& i = m& p + m& s = ρApVp + ρAsVs
m& e = ρAeVe
∴ApVp + AsVs = AeVe
So
e
p p s s
e A
A V A V
V
+
=
Ae = As + Ap
2p
p D
4
A
π
= 2p
2
s o D
4
D
4
A
π
−
π
= 2
e Do
4
A
π
=
i e
s
p
3
( ) 2 ( p s )
o
2p
2 s
o
s
2p
2
p o
2p
e
p p s s
e V V
D
D
V
D
D V D D V
A
A V A V
V = + −
+ −
=
+
=
(20 5) 6.6667m/ s
30
5 102
2
= + − =
Problem 4. – A slow leak develops in a storage bottle and oxygen slowly leaks out. The volume
of the bottle is 0.1 m3 and the diameter of the hole is 0.1 mm. The initial pressure is 10 MPa and
the temperature is 20°C. The oxygen escapes through the hole according to the relation
e Ae
T
m& = 0.04248 p
where p is the tank pressure and T is the tank temperature. The constant 0.04248 is based on the
gas constant and the ratio of specific heats of oxygen. The units are: pressure N/m2, temperature
K, area m2 and mass flow rate kg/s. Assuming that the temperature of the oxygen in the bottle
does not change with time, determine the time it takes to reduce the pressure to one half of its
initial value.
∀ = 0.1 m3
p1 = 10 MPa
T1 = 293K = T2
p2 = 5 MPa
kg K
259.8219 J
32
R 8,314.3
⋅
= =
From the continuity equation
me
dt
dm = − &
but
RT
m p
∀
=
so
p
T
0.04248 A
m
dt
dp
dt RT
dm e
= − e = −
∀
= &
Integrating we get,
O2
m(t)
d = 0.1mm
e Ae
T
m& = 0.04248 p
4
t
0.04248 R TA
p
p
ln e
1
2
⎟ ⎟
⎠
⎞
⎜ ⎜
⎝
⎛
∀
= −
46,713.4076sec 12.9759 hrs
2
ln 1
(259.8219) 293
1000 mm
0.1 mm m
4
(0.04248)
0.1
p
p
ln
(0.04248)A R T
t
2
1
2
e
= =
⎟⎠
⎞
⎜⎝
⎛
⎟⎠
⎞
⎜⎝
⎛
⎟⎠
⎞
⎜⎝
⎛ π
= −
∀
= −
Problem 5. – A normal shock wave occurs in a nozzle in which air is steadily flowing. Because
the shock has a very small thickness, changes in flow variables across the shock may be assumed
to occur without change of cross-sectional area. The velocity just upstream of the shock is 500
m/s, the static pressure is 50 kPa and the static temperature is 250 K. On the downstream side of
the shock the pressure is 137 kPa and the temperature is 343.3 K. Determine the velocity of the
air just downstream of the shock.
V1 = 500 m/ s V2 = ?
p1 = 50 kPa p2 = 137 kPa
T1 = 250 K T2 = 343.3 K
A1 = A2
From the continuity equation
m& 1 = m& 2
So
ρ1A1V1 = ρ2A2V2
(500) 250.5839m/ s
250
343.3
137
V 50
T
T
p
V p
p / RT
V V p / RT 1
1
2
2
1
1
2 2
1 1
1
2
1
2 = ⎟⎠
⎞
⎜⎝
⎛
⎟⎠
⎞
⎜⎝
= = = ⎛
ρ
ρ
=
2
1
5
Problem 6. – A gas flows steadily in a 2.0 cm diameter circular tube with a uniform velocity of
1.0 cm/s and a density ρo. At a cross section farther down the tube, the velocity distribution is
given by V = Uo[1-( r/R)2], with r in centimeters. Find Uo, assuming the gas density to be
ρo[1+( r/R)2].
V1 = 1 cm/ s
⎥ ⎥
⎦
⎤
⎢ ⎢
⎣
⎡
⎟⎠
⎞
⎜⎝
= − ⎛
2
2 o R
V U 1 r
ρ1 = ρo
⎥ ⎥
⎦
⎤
⎢ ⎢
⎣
⎡
⎟⎠
⎞
⎜⎝
ρ = ρ + ⎛
2
2 o R
1 r
m& 1 = m& 2
o
2 2
o 1
R
1 o 1 1
R
1 o 1 m& = ∫ ρ V dA = ∫ ρ V 2πrdr = ρ V πR = πR ρ
( )
o o
2 2
o o
1
o
2 2 5
o o
2
2
2 o
R 2
o
R
2 2 2 o o
R U
3
2
6
1
2
2 U R 1
R
U 2 R d where r
2 rdr
R
U 1 r
R
m V dA 1 r
ρ π = ⎟⎠
⎞
⎜⎝
= πρ ⎛ −
= ρ π ξ − ξ ξ ξ =
π ⎟
⎟
⎠
⎞
⎜ ⎜
⎝
⎛
− ⎟
⎟
⎠
⎞
⎜ ⎜
⎝
⎛
= ρ = ρ +
∫
& ∫ ∫
o o
2
o
2 R U
3
R 2 ρ
π
∴π ρ =
so cm/ s
2
V 3 o =
Problem 7. – For the rocket shown in Figure 1.6, determine the thrust. Assume that exit plane
pressure is equal to ambient pressure.
( ) ( ) ( )
e e
2
H o
e e
H o
e atm e e e H o A
m m
V
m m
p p A m V 0 m m
ρ
+
= ⎟
⎟⎠
⎞
⎜ ⎜⎝
⎛
ρ
+
= − + = + +
& & & &
T & & &
r
1 2
6
Problem 8. – Determine the force F required to push the flat plate of Figure Pl.8 against the
round air jet with a velocity of 10 cm/s. The air jet velocity is 100 cm/s, with a jet diameter of 5.0
cm. Air density is 1.2 kg/m3.
Figure P1.8
To obtain steady state add + Vp to all velocities
F = m& V
( ) (0.5) (1 0.1)
4
2 . 1 AV m 2 + ⎟⎠
⎞
⎜⎝
⎛ π
& = ρ = = 0. kg / s
F = (0.)(1.1) = 0. N
Problem 9. – A jet engine (Figure P1.9) is traveling through the air with a forward velocity of
300 m/s. The exhaust gases leave the nozzle with an exit velocity of 800 m/s with respect to the
nozzle. If the mass flow rate through the engine is 10 kg/s, determine the jet engine thrust. Exit
plane static pressure is 80 kPa, inlet plane static pressure is 20 kPa, ambient pressure surrounding
the engine is 20 kPa, and the exit plane area is 4.0 m2.
F
x
V = -10 s
cm
Vj = 100 s
cm
Vj = 110 s
cm
x
V = 0
F
7
Figure P1.9
T = (pe − patm )Ae + m& (Ve − Vi ) = (80 − 20)(4)+ (10)(800 − 300) = 240 + 5 = 245kN
Problem 10. – A high-pressure oxygen cylinder, typically found in most welding shops,
accidentally is knocked over and the valve on top of the cylinder breaks off. This creates a hole
with a cross-sectional area of 6.5 x 10-4 m2. Prior to the accident, the internal pressure of the
oxygen is 14 MPa and the temperature is 27°C. Based on critical flow calculations, the velocity
of the oxygen exiting the cylinder is estimated to be 300 m/s, the exit pressure 7.4 MPa and the
exit temperature 250 K. How much thrust does the oxygen being expelled from the cylinder
generate? What percentage is due to the pressure difference? What percentage due to the exiting
momentum? Atmospheric pressure is 101 kPa. Also note that 0.2248 lbf = 1 N.
Figure P1.10
Ve = 300 m/ s 4 2
Ae 6.5 10 m = × −
pe = 7.4 MPa patm = 101 kPa = 0.101 MPa
Te = 250 k e e
e
e
e e e A V
RT
p
m& = ρ A V =
kg k
R 259.82 J
⋅
= ( )( ) (6.5 10 )(300) 22.2 kg / s
259.82 250
m
7.4 10 N
2 4
6
× =
⎟ ⎟⎠
⎞
⎜ ⎜⎝
⎛
×
= −
300 m/s 800 m/s
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